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At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then
- a)potential at its surface is 150V.
- b)the charge on the sphere is (5/3) × 10-10C.
- c)the electric field on the surface is 1500 V/m.
- d)the electric potential at its centre is 225V.
Correct answer is option 'A,C,D'. Can you explain this answer?
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We can use the formula for potential due to a uniformly charged sphere:
V = (1/4πε₀) (Q/r)
where V is the potential, Q is the charge on the sphere, r is the distance from the center of the sphere, and ε₀ is the permittivity of free space.
At a distance of 5 cm, we have:
100 = (1/4πε₀) (Q/0.05)
At a distance of 10 cm, we have:
75 = (1/4πε₀) (Q/0.1)
Dividing the two equations, we get:
100/75 = (0.1/0.05)
Simplifying, we get:
4/3 = Q/0.05Q
Solving for Q, we get:
Q = (4/3)(0.05Q) = 0.0667Q
So the charge on the sphere is:
Q = (3/4)(0.0667) = 0.05 C
Now we can use the potential at the surface to find the total charge:
150 = (1/4πε₀) (Q/0.1)
Solving for Q, we get:
Q = 0.15 C
So the answer is:
a) The potential at the surface is 150V.
b) The charge on the sphere is 0.15 C, which is not equal to (5/3) C.
V = (1/4πε₀) (Q/r)
where V is the potential, Q is the charge on the sphere, r is the distance from the center of the sphere, and ε₀ is the permittivity of free space.
At a distance of 5 cm, we have:
100 = (1/4πε₀) (Q/0.05)
At a distance of 10 cm, we have:
75 = (1/4πε₀) (Q/0.1)
Dividing the two equations, we get:
100/75 = (0.1/0.05)
Simplifying, we get:
4/3 = Q/0.05Q
Solving for Q, we get:
Q = (4/3)(0.05Q) = 0.0667Q
So the charge on the sphere is:
Q = (3/4)(0.0667) = 0.05 C
Now we can use the potential at the surface to find the total charge:
150 = (1/4πε₀) (Q/0.1)
Solving for Q, we get:
Q = 0.15 C
So the answer is:
a) The potential at the surface is 150V.
b) The charge on the sphere is 0.15 C, which is not equal to (5/3) C.
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At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then a)potential at its surface is 150V. b)the charge on the sphere is (5/3) × 10-10C.c)the electric field on the surface is 1500 V/m.d)the electric potential at its centre is 225V.Correct answer is option 'A,C,D'. Can you explain this answer?
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At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then a)potential at its surface is 150V. b)the charge on the sphere is (5/3) × 10-10C.c)the electric field on the surface is 1500 V/m.d)the electric potential at its centre is 225V.Correct answer is option 'A,C,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then a)potential at its surface is 150V. b)the charge on the sphere is (5/3) × 10-10C.c)the electric field on the surface is 1500 V/m.d)the electric potential at its centre is 225V.Correct answer is option 'A,C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then a)potential at its surface is 150V. b)the charge on the sphere is (5/3) × 10-10C.c)the electric field on the surface is 1500 V/m.d)the electric potential at its centre is 225V.Correct answer is option 'A,C,D'. Can you explain this answer?.
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